Problem: The polynomial $p(x)=3x^3-5x^2-4x+4$ has a known factor of $(x-2)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
Solution: We know $(x-2)$ is a factor of $p(x)$. This means that $p(x)=(x-2)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x-2)$ $\begin{array}{r} 3x^2+\phantom{4}x-2 \\ x-2|\overline{3x^3-5x^2-4x+4} \\ \mathllap{-(}\underline{3x^3-6x^2\phantom{-4x+4}\rlap )} \\ x^2-4x+4 \\ \mathllap{-(}\underline{x^2-2x\phantom{+4}\rlap )} \\ -2x+4 \\ \mathllap{-(}\underline{-2x+4\rlap )} \\ 0 \end{array}$ We find that $q(x)=3x^2+x-2$. Factoring $q(x)$ We can factor $q(x)$ by grouping: $\begin{aligned} q(x)&=3x^2+x-2 \\\\ &=3x^2+3x-2x-2 \\\\ &=3x(x+1)-2(x+1) \\\\ &=(3x-2)(x+1) \end{aligned}$ Putting it all together $\begin{aligned} p(x)&=3x^3-5x^2-4x+4 \\\\ &=(x-2)(3x^2+x-2) \\\\ &=(x-2)(3x-2)(x+1) \end{aligned}$